91Ó°ÊÓ

In spectrophotometry, light absorption is a key concept. It involves measuring how much light a sample absorbs when light passes through it. Light absorption is used to determine the concentration of a substance in a solution. The general idea is that the more concentrated the solution, the more light it will absorb. Depending on the wavelength of light used, different substances will absorb varying amounts of light.

For instance, in the given exercise, we use the spectrophotometer to shine light through a solution and measure the light's intensity before ()(\(I_0\)) and after () \(I\)) passing through the sample. Absorption is calculated using the ratio \(\frac{I}{I_0}\). This ratio tells us the percentage of light that makes it through the solution as opposed to being absorbed by it.

Calculating the concentration of a substance involves using a known relationship between light absorption and concentration. This relationship is often modeled mathematically using a formula. In this exercise, the given formula is:
\[C = -2500 \ln \left(\frac{I}{I_{0}}\right)\]

• First, identify the ratio of the intensity of light that emerges \( I \) to the intensity of the incident light \( I_0 \). Here, it is provided that \( \frac{I}{I_0} = 0.70 \).
• Plug this ratio into the formula for concentration. It now becomes:
  \[C = -2500 \ln(0.70)\]
• The next step involves computing the natural logarithm of \(0.70\), which brings us to the mathematical concept of the natural logarithm.

The natural logarithm, often abbreviated as "ln," is a mathematical function that calculates the logarithm to the base \(e\), where \(e\) is approximately 2.718. It is particularly useful in exponential growth or decay processes and is often encountered in chemical kinetics, like spectrophotometry.

Computing \( \ln(0.70) \) tells us the power to which \( e \) must be raised to get 0.70. In this exercise, \( \ln(0.70) \approx -0.3567 \). This value is then multiplied by the constant \(-2500\), reflecting how absorption relates to concentration. Hence:
\[C = -2500 \times (-0.3567) = 891.75\]

Using the natural logarithm allows us to handle ratios like \( \frac{I}{I_0} \) effectively, converting them into measures reflecting the concentration of a substance due to its exponential nature.